The mechanics of friction and compliance in multi-contact
arrangements is key to understanding and predicting grasp
stability and dynamic response to external loads.

{\em force closure} is a generic property of frictional
equilibrium grasps~\cite{nguyen,yoshikawa}\footnote{Force closure
requires that the contacts be able to balance any disturbance
wrench (force and torque) acting on the grasped
object~\cite{bicchi_survey,salisbury82}.}, which ensures that the
grasp is locally controllable~\cite{brook&shoham,nguyen}.

Friction effects play a key role in virtually all
light-to-moderate duty grasping and manipulation applications. For
instance, friction effects allow stable grasping with a much
smaller number of contacts than would otherwise be required---two
contacts rather than four in {\small 2D}, and two or three
contacts rather than seven in {\small
3D}~cite{brook&shoham,nguyen,bicchi_survey}.

The workpiece need not only be stable against external
perturbations, but it must also stay within a specified tolerance
in response to external loads.

, such as the Hertz model \cite{hertz} and Walton model
\cite{Walton1978},

Other lumped parameter compliant models include the work of Elata
\cite{elata&berryman} which enumerates several other lumped
parameter models, of a greater complexity then that of the Walton
model considered in this paper.

Rigid-body models are also available which can predict both normal
and tangential forces. These models are even simpler then lumped
parameters models in the sense that they require no state to model
the contact, however typically rigid-body models suffer from both
ambiguity and inconsistency. A survey of papers in this extensive
area is provided in the paper by Stewart \cite{Stewart2000}. Or
and Rimon \cite{Or2005} and Greenfield Rizzi and Choset
\cite{Greenfield2005} made a lot of progress in analyzing the
possible set of contact forces and possible loci of equilibria
configurations. However under the rigid body assumption if a small
perturbation from equilibrium is made, it is very hard to predict
the direction of the flow of the robot's dynamic system. The
reason for that is as follows: when a small perturbation is made,
a contact may be broken. Then the system that govern the dynamics
of the robot is switched to a different system that includes one
less contact point. Then after the dynamics further evolves a new
contact may be made or another contact may break, and the dynamics
switches again to a different system. It is very hard to predict
the correct switching sequence for a given perturbation.
Therefore, to the best of our knowledge there is no general
stability criterion and proof for rigid object that contact a
rigid environment. However, a stability proof does exist under the
assumption that contacts do not break
\cite{Arimoto2005,Cheah1998}. Obviously that is not the general
case when dealing with on-line control of grasping and fixturing
arrangements.

The first widely accepted contact model in the solid mechanics
literature is the normal compliance model of Hertz
\cite{Hertz1882}[1882]. Hertz restricts his attention to the
normal loading of two elastic solids under the assumption that the
area of contact is small in comparison to the radii of curvature
and size of the contacting bodies.

In the solid mechanics literature, The first focused effort on
formulating analytical friction-compliance models is by Mindlin
and Deresiewicz \cite{mindlin49,mindlin53}[1949-1953]. They
investigate the case where a contact is initially loaded along the
normal direction in accordance with the Hertz normal compliance
model. Then they analyze the tangential traction field generated
by applying pure tangential loads while the normal penetration
remains constant. Their investigation revealed highly nonlinear
and complex phenomena such as micro-slip and hysteresis
\cite{johnson}. Motivated by granular material packing
applications, Walton
\cite{Walton1978,walton87,elata&berryma}[1987] derived an
analytical friction-compliance model which is more relevant for
grasping applications. Walton considers the oblique loading of two
identical spheres under the assumption that the spheres are loaded
with a constant ratio of normal to tangential displacements, such
that no slip occurs during loading.  As with Hertz, the work of
Walton assumes small displacements. Our ensuing results are based
on Walton's tangential compliance model.

%Although, the normal loading assumption is overly restrictive for
%the contact conditions we wish to consider, the resultant normal
%force-displacement law agrees with the normal force-displacement
%component of other contact models which also include a tangential
%traction component, including the one considered in this paper.

%In practice, the physics of contact is difficult to model,
%particularly with computationally tractable lumped parameter
%models.  Lumped parameter models in the solid mechanics literature
%attempt to model the contact forces using a small number of state
%variables like normal and tangential displacements, in order to
%provide simplicity.
%However, due to the many complicating factors in
%predicting contact force, such as hysteresis, velocity dependence,
%energy loss, creep, and microslip, these simple models invariably
%\emph{sacrifice accuracy for simplicity}.

%and that certain energy flow requirements are satisfied. The last
%is a necessary condition in order to assure uniqueness of solution
%to the contact equations.

%A more recent paper by Elata
%\cite{Elata1996a} critiques this work by pointing out the
%thermodynamic inconsistency of this model and noting that there
%exist closed loop paths which generate energy. Despite the obvious
%physical impossibility of this phenomenon, the Walton model
%remains attractive due to its simplicity.

Two more works that deals with frictional contact model and worth
mention are the work by Xiong et al. \cite{Xiong2005}, that
analyzed grasp stability and implemented a contact model that was
developed by Sinha and Abel \cite{Sinha1992}. This contact model
assumes linear elasticity and discretizes the continuous pressure
distribution. While this contact model is very similar to the
Walton model it does not take into account the dependency on
loading path, which is critical according to the solid mechanics
literature~\cite{johnson}. Moreover, in this work the authors use
a potential function approach to analyze the grasp stability. In
contrast, the Hertz-Walton model contains inherent asymmetry due
to the dependence of the normal contact force on the normal
deflection while the tangential force depends on {\em both the
tangential and the normal deflections}. This requires us to use a
different tool to analyze the grasp stability. We recently
developed a stability criterion for asymmetric linear systems
\cite{ShapiroAsymmetry2005}. We will use this new tool to develop
grasp stability criteria.

with all its parameters being material coefficients and geometric
parameters of the contacting bodies.

natural compliance at a frictional contact and its effect on the
stiffness and stability of grasp and fixture arrangements. The
paper introduces

A related example is whole arm manipulation where certain mid-link
contacts can only passively react to external loads~\cite{omata}.
The synthesis of whole arm grasps that can passively resist
external loads requires a model for the reaction forces arising
from natural friction and compliance effects.

The main objective of this paper is to integrate friction and
compliance contact phenomena into a single law. The law should be
based on the solid mechanics literature, and be formulated in an
analytic lumped parameter form suitable for on-line grasping
applications.

%Ideally, a contact model should be both accurate enough to capture
%all the relevant physics of interaction, broad enough to apply in
%a wide range of contact conditions, and simple enough so that
%control decisions can tractably be made on the basis of the
%contact model.

%All this work is based on assumptions which may not hold true in
%practice due to finite displacements, or other violations of
%modeling assumptions.  It is therefore useful to
%\emph{experimentally} validate the use of these contact models.
%The work of Burdick et. al. \cite{Burdick2003} describes a high
%fidelity experimental setup which is used to test the validity of
%the Hertz model for fixturing applications. This paper seeks to
%experimentally determine the applicability of a parametric model
%based on the

giving the
linearized relation between small object movement and the contact
reaction force. These matrices
The determine the composite stiffness
matrix of the entire grasp or fixture.

, called the {\em
contact stiffness matrix.} A key result is that each of
Hence grasp stability must be determined by the linearized
dynamics of the contact arrangement about a candidate equilibrium
grasp, rather than being a local-minimum test of some potential
energy function. Section~\ref{sec.stiffness} also characterizes
the

 and press against
$\Bs$ in static equilibrium. (The ensuing contact models
generalize to arbitrarily shaped fingertips as long as they
initially maintain point contact with $\Bs$.)

This approach offers excellent accuracy, but compliance models
relating contact reaction force to contact displacement can only
be obtained numerically.


Next we will discuss the derivative of the terms that include the
contact forces with respect to $p_1$. An explicit formula for $f_i
= R_i \mbox{\scriptsize $ \vctwo{f_i^t}{f_i^n}$}$ is obtained by
substituting for

%We can write $f_i(p_1,p_2)= R_i(p_1)
%\tilde{f}_i(\vect{\Dg}_i(p_1),\dot{\vect{\Dg}}_i(p_1,p_2))$. Thus,

\beq{K_p} \barr & \frac{\partial}{\partial p_1}G_i^T(p_1)
R_i(p_1)\tilde{f}_i(p_1,p_2) =\\ &D^2 X_{r_i}(p_1)R_i \tilde{f}_i
+G_i^T DR_i(p_1)\tilde{f}_i +G_i^T R_i \big\{
\frac{\partial}{\partial \vect{\Dg}_i} \tilde{f}_i
D\vect{\Dg}_i(p_1)+ \frac{\partial}{\partial \dot{\vect{\Dg}}_i}
\tilde{f}_i \frac{\partial}{\partial p_1 }\dot{\vect{\Dg}}_i
\big\}. \earr
 \eeq

The $i^{th}$ contact force is given by  $f_i = R_i
\mbox{\scriptsize $ \vctwo{f_i^t}{f_i^n}$}$. f_i(\mbox{\small
$\vect{\Dg}_i,\dot{\vect{\Dg}}_i$} )

R_i \mbox{\scriptsize $ \vctwo{f_i^t}{f_i^n}$}

\tilde{f}_i D\vect{\Dg}_i(p_1)+ \frac{\partial}{\partial
\dot{\vect{\Dg}}_i} \tilde{f}_i \frac{\partial}{\partial p_1
}\dot{\vect{\Dg}}_i

\frac{\partial}{\partial \vect{\Dg}_i} \tilde{f}_i= K_i+
\frac{\partial}{\partial \vect{\Dg}_i}(\varphi_i^t,\varphi_i^n)$,

%--------------------------------here 6/7/06-------------------------
Previously we showed that $\frac{\partial}{\partial p_1}
\tilde{f}_i$ contains a term which corresponds to the derivative
of the dissipative functions $(\varphi_i^t,\varphi_i^n)$ with
respect to $\vect{\Dg}_i$. At the equilibrium state $(q_0,0)$ we
have that $\dot{\vect{\delta}}_i=0$. Hence
$\frac{\partial}{\partial \vect{\delta}}_i
(\varphi_i^t,\varphi_i^n)$ vanishes at the equilibrium.

Using \eq{nu}, $\frac{d}{dt} \Dg^t_i(q(t))= t_i\cdot G_i \qd$.
Hence $D\Dg^t_i(p_1) = t_i^T G_i(p_1)$. Similarly, it can be
verified that the time derivative of $\Dg_i^n(q(t)$ satisfies
$\frac{d}{dt} \Dg^n_i(q(t))= -n_i\cdot G_i
\qd$~\cite{indet_icra03}. Hence $D\Dg^n_i(p_1) = -n_i^T G_i(p_1)$.
Collecting the two differentials together, we obtain
$D\vect{\Dg}_i(p_1)= -R_i^T G_i(p_1)$ where $R_i=[t_i \, n_i]$.

since its core consists of the asymmetric contact stiffness
matrices $K_i(q_0,0)$ for $i=1\ldots k$

$\mbox{\scriptsize $\left( \!\!\begin{array}{c} \DG \Dg_i^t\\
\DG \Dg_i^n
\end{array} \!\!\right)$} = \DG \vect{\Dg}_i(q_0)
\DG q$

Under these assumptions The $i^{th}$ contact force acting on $\Bs$
is $f_i(q,\qd)= R_i \mbox{\scriptsize $\left( \!\begin{array}{c}
f_i^t
\\ f_i^n \end{array} \!\right)$}$, where $(f_i^t,f_i^n)$ are specified
by the Hertz-Walton laws.

-----------------------------------


%Consider the second order linear asymmetric system:
%
%\vspace{-.0in}
%\begin{equation}
%\ddot{p}+K_d \dot{p} + K_p p =0,
% \label{eq.simple_linear_sys}
%\end{equation}
%
%\vspace{-.00in} \noindent where $K_d \in \real^{n \times n}$ is
%symmetric positive definite, $K_p \in \real^{n \times n}$ is
%asymmetric, while its symmetric part $(K_p)_s$ is positive
%definite. The following theorem states that if the skew-symmetric
%part of $K_p$, $(K_i)_{as}$, is sufficiently small, the system
%\eq{simple_linear_sys} is globally asymptotically stable.
%
%\begin{thm}[global asymptotic stability] Consider
%the system of (\ref{eq.simple_linear_sys}). Let $\beta \in \real$
%be the minimal eigenvalue of $K_d $. Let $\alpha \in \real$ be the
%minimal eigenvalue of $(K_p)_s$, and let $\Gg \in \real$ be the
%matrix norm\footnote{The matrix norm is defined as
%$\|E\|=max\{\|Eu\|\}$ over all vectors $\|u\| \leq 1$.} of the
%skew-symmetric part of $K_p$. Then if
%
%\vspace{-.0in}
%\[
%|\Gg| < \sqrt{\alpha} \beta
%\]
%
%
%\vspace{-.00in}\noindent the system \eq{simple_linear_sys} is
%globally asymptotically stable.
%\label{thm.stability_linear_asymetric_sys}
%\end{thm}
%
%\noindent \textbf{Proof:} The system (\ref{eq.simple_linear_sys})
%can be written as
%
%\vspace{-.0in}
%\[
%\frac{d}{dt} \vctwo{p}{\dot{p}} =
%\underbrace{\matwo{0}{I}{-K_p}{-K_d}}_{A} \vctwo{p}{\dot{p}}
%\]
%
%
%\vspace{-.0in}\noindent For global asymptotic stability, it
%suffices to show that the real part of the eigenvalues of $A$ is
%negative. Let $\lambda \in \complex$ be an eigenvalue of $A$ with
%corresponding eigenvector $v=(v_1,v_2)\in \complex^{2n}$ ($v \neq
%0$). Note that each $v_i$ is a complex vector in $\complex^n$.
%Then
%
%\vspace{-.0in}
%\[ \barr
%& \hspace{-.4in} \matwo{0}{I}{-K_p}{-K_d} \vctwo{v_1}{v_2}  \\
%& = \vctwo{v_2}{-K_p v_1 - K_d v_2} = \lambda \vctwo{v_1}{v_2}.
%\earr
%\]
%
%\vspace{-.0in}\noindent Clearly, $\lambda \!=\! 0$ cannot be an
%eigenvalue of $A$. Since $\lambda \neq 0$, it follows that $v_1
%\neq \vec{0}$ and $v_2 \neq \vec{0}$. Hence we may assume without
%loss of generality that $v_1^* \cdot v_1 =1$, where $*$ denotes
%complex conjugate transpose. Based on this choice, we can write
%$\lambda^2 = v_1^* \lambda^2 v_1 = v_1^* \lambda v_2 = v_1^* (-K_p
%v_1 -K_d v_2)= -v_1^* K_p v_1 - \lambda v_1^* K_d v_1$,
%%\begin{equation}
%%\begin{array}{lcl}
%%\lambda^2 &=& v_1^* \lambda^2 v_1 = v_1^* \lambda v_2 = v_1^*
%% &=& -v_1^* K_p v_1 - \lambda v_1^* K_d v_1,
%% \end{array}
%%\label{eq.lambda_square}
%%\end{equation}
%where we used the relations $\Lg v_1=v_2$ and $\Lg v_2= -K_p v_1
%-K_d v_2$. Since $K_d >0$, the scalar $\tilde{\beta}=v_1^* K_d
%v_1$ is positive real. Similarly, the scalar $\tilde{\alpha}=
%v_1^* (K_p)_s v_1$ is also positive real. Since $(K_p)_{as}$ is
%skew-symmetric, we can write $j\tilde{\Gg}= v_1^* (K_p)_{as} v_1$,
%where $j=\mbox{\small $ \sqrt{-1}$}$ and $\tilde{\Gg}$ is real.
%Substituting these scalars into the quadratic equation of $\Lg$
%gives
%
%\vspace{-.0in}
%\begin{equation}
%\lambda^2 + \tilde{\beta} \lambda + \tilde{\alpha} + j
%\tilde{\Gg}=0. \label{eq.lambda_square_equation}
%\end{equation}
%
%
%\vspace{-.00in}\noindent Note that every eigenvalue of $A$
%satisfies this equation. The solution of
%\eq{lambda_square_equation} is:
%
%\vspace{-.0in}
%\begin{equation}
%\lambda_{1,2}=\half \left(-\tilde{\beta} \pm
%\sqrt{\tilde{\beta}^2- 4 (\tilde{\alpha} + j \tilde{\Gg})}
%\right). \label{eq.lambda_solutions}
%\end{equation}
%
%
%\vspace{-.0in}\noindent Let us pause to recall how one computes
%the square root of a complex number. Consider a complex number
%$z=a+ jb$ with a norm $\absvl{z}=\sqrt{a^2+b^2}$ and argument
%$\theta=\arctan(b/a)$. Then $\sqrt{z}=\pm (a^2+b^2)^{\frac{1}{4}}
%\angle \frac{\theta}{2}$, and in cartesian coordinates
%$\sqrt{z}=\pm (a^2+b^2)^{\frac{1}{4}} \big( \cos
%\left(\frac{\theta}{2} \big) + j \sin \big(\frac{\theta}{2} \big)
%\right)$. Since $\cos(\theta)=\frac{a}{\sqrt{a^2+b^2}}$, we use
%the trigonometric identity $\cos \big( \frac{\theta}{2} \big) =
%\sqrt{\frac{1+ \cos ( \theta )}{2}} $ to obtain
%
%\vspace{-.0in}
%\[
%\re{ \sqrt{z} } = \pm \frac{(a^2+b^2)^{\frac{1}{4}}}{\mbox{\small
%$\sqrt{2}$}} \left(1 + \frac{a}{\sqrt{a^2+b^2}} \right) ^{\half}.
%\]
%
%
%\vspace{-.0in}\noindent In our case $a=\tilde{\beta}^2 - 4
%\tilde{\alpha}$ and $b= -4 \tilde{\Gg}$, and \eq{lambda_solutions}
%implies that
%
%\vspace{-.0in}
%\[ \barr
%& \re{\lambda_{1,2}} =
%-\frac{\tilde{\beta}}{\mbox{\small $2$}} \\
%& \pm \frac{\big( (\tilde{\beta}^2 - 4 \tilde{\alpha})^2+16
%\tilde{\Gg}^2 \big)^{\frac{1}{4}}}{\mbox{\small $2\sqrt{2}$}}
%\big( 1 + \frac{(\tilde{\beta}^2 - 4
%\tilde{\alpha})}{\sqrt{(\tilde{\beta}^2 - 4 \tilde{\alpha})^2+16
%\tilde{\Gg}^2}} \big) ^{\half} \earr
%\]
%
%\vspace{-.0in}\noindent The requirement $\re{\lambda_{1,2}}<0$
%introduces an inequality in $\tilde{\alpha}$, $\tilde{\beta}$, and
%$\tilde{\Gg}$. Rearranging terms in this inequality gives the
%equivalent inequality,
%%\[
%%\tilde{\beta}>\left( (\tilde{\beta}^2 - 4 \tilde{\alpha})^2+16
%%\tilde{\Gg}^2 \right)^{\frac{1}{4}} \left( \half + \half
%%\frac{(\tilde{\beta}^2 - 4 \tilde{\alpha})}{\sqrt{(\tilde{\beta}^2
%%- 4 \tilde{\alpha})^2+16 \tilde{\Gg}^2}} \right) ^{\half}.
%%\]
%%Taking the square of both sides and then multiply the inequality
%%by $2$ gives
%%\[
%%2\tilde{\beta}^2>\left( (\tilde{\beta}^2 - 4 \tilde{\alpha})^2+16
%%\tilde{\Gg}^2 \right)^{\half} \left(1+ \frac{(\tilde{\beta}^2 -
%%4 \tilde{\alpha})}{\sqrt{(\tilde{\beta}^2 - 4 \tilde{\alpha})^2+16
%%\tilde{\Gg}^2}} \right).
%%\]
%%Rearranging this inequality results in
%%\[
%%2\tilde{\beta}^2>\left( (\tilde{\beta}^2 - 4 \tilde{\alpha})^2+16
%%\tilde{\Gg}^2 \right)^{\half} + \tilde{\beta}^2 - 4
%%\tilde{\alpha}.
%%\]
%%Next we add $4 \tilde{\alpha} - \tilde{\beta}^2$ to both sides and
%%then take the square of both sides to have
%
%\vspace{-.0in}
%\[
%\left(4
%\tilde{\alpha}+\tilde{\beta}^2\right)^2>\left(\tilde{\beta}^2 - 4
%\tilde{\alpha}\right)^2+16 \tilde{\Gg}^2.
%\]
%
%\vspace{-.0in}\noindent Cancelling similar terms  yields the
%inequality
%%Opening the brackets gives
%%\[
%%16 \tilde{\alpha}^2 + \tilde{\beta}^4 + 8\tilde{\alpha}
%%\tilde{\beta}^2 > \tilde{\beta}^4 + 16 \tilde{\alpha}^2
%%-8\tilde{\alpha} \tilde{\beta}^2 +16 \tilde{\Gg}^2.
%%\]
%%Rearranging this and dividing by $16$ results in
%%\[
%%\tilde{\alpha} \tilde{\beta}^2 >\tilde{\Gg}^2,
%%\]
%%or simply
%
%\vspace{-.0in} \beq{must} \absvl{\tilde{\Gg}} <
%\sqrt{\tilde{\alpha}} \tilde{\beta}.
% \eeq
%%Recall now that $\tilde{\beta} = v_1^* K_d v_1 > 0$,
%%$\tilde{\alpha}= v_1^* (K_p)_s v_1 > 0$, and $i\tilde{\Gg}= v_1^*
%%(K_p)_{as} v_1$.
%
%\vspace{-.0in}\noindent For stability we must ensure that the
%inequality \eq{must} holds for every $\tilde{\alpha}$,
%$\tilde{\beta}$, and $\tilde{\Gg}$. In other words, \eq{must} must
%hold for every eigenvalue $\lambda$ and associated eigenvector $v$
%of $A$. Therefore we bound $\tilde{\alpha}$, $\tilde{\beta}$, and
%$\tilde{\Gg}$ as follows.
%%=\lambda_{min}\left((K_p)_s\right) \|v_1\|^2
%First, $0<\alpha = \lambda_{min}\left((K_p)_s\right)\leq v_1^*
%(K_p)_s v_1=\tilde{\alpha}$. Second,
%%=\lambda_{min}\left(K_d\right) \|v_1\|^2
%$0<\beta = \lambda_{min}\left(K_d\right)\leq v_1^* K_d
%v_1=\tilde{\beta}$.
%%=\|v_1^*\| \|(K_p)_{as}\| \|v_1\|
%Third, $\absvl{\Gg} = \|(K_p)_{as}\| \geq |v_1^* (K_p)_{as} v_1| =
%\absvl{j \tilde{\Gg}}= \absvl{\tilde{\Gg}}$. Using these bounds,
%$\Gg < \sqrt{\alpha} \beta$ implies that
%$\absvl{\tilde{\Gg}}<\sqrt{\tilde{\alpha}} \tilde{\beta}$ for
%every $\tilde{\alpha}$, $\tilde{\beta}$, and $\tilde{\Gg}$. \epf

%%---------------------------------------------------------------
%\vspace{.00in} \noindent {\bf Application to grasp stability.} The
%linearized grasp system contains an additional term, the inertia
%matrix $M(q_0)$ which multiplies $\ddot{p}$.

---------------------------------------

Next we show when these conditions can be satisfied by a grasping
arrangement.
%%-------first condition-------------------------------------
To satisfy the \textbf{first condition} we need:
\[
\lambda_{min}\left(M^{-1/2} \sum_{i=1}^k  J_i^T R_i
\matwo{\frac{\partial \varphi_i^t}{\partial
\dot{\delta}_i^t}}{0}{0}{\frac{\partial \varphi_i^n}{\partial
\dot{\delta}_i^n}} R_i^T J_i M^{-1/2}\right)>0.
\]
In our case $\matwo{\frac{\partial \varphi_i^t}{\partial
\dot{\delta}_i^t}}{0}{0}{\frac{\partial \varphi_i^n}{\partial
\dot{\delta}_i^n}}$ is symmetric positive definite matrix.
Consequently the matrix $\sum_{i=1}^k  J_i^T R_i
\matwo{\frac{\partial \varphi_i^t}{\partial
\dot{\delta}_i^t}}{0}{0}{\frac{\partial \varphi_i^n}{\partial
\dot{\delta}_i^n}} R_i^T J_i$ is also positive definite since it
is a sum of quadratic forms of positive definite matrices.

The following lemma assert that the transformation $M^{-1/2} K_d
M^{-1/2}$ does not change the positive definiteness property of
$K_d$.

\blemm{symmetric_quad_trans} Let $M \in \real^{n \times n}$ be a
symmetric positive definite matrix, and let $A \in \real^{n \times
n}$ be a symmetric matrix. Thus if $A$ is positive definite
$M^{-1/2} A M^{-1/2}$ is positive definite. \elemm

\bpf If $A$ is positive definite then
\begin{equation}
v^T A v >0 \mbox{  } \forall \vec{v} \neq 0. \label{eq.pos_A}
\end{equation}
And we need to show that
\begin{equation}
v^T M^{-1/2} A M^{-1/2} v >0 \mbox{  } \forall v \neq 0.
\label{eq.pos_trans_A}
\end{equation}
%Note that since $M^{-1/2}$ is unique it is also symmetric.
Note that  $M^{-1/2}$ is non-singular. Therefore we can assign
$\vec{u}=M^{-1/2} v$ and rewrite (\ref{eq.pos_trans_A}) as
\[
u^T A u,
\]
but from (\ref{eq.pos_A}) we know this term is positive for
$\vec{u},\vec{v} \neq 0$. \epf Hence, we conclude that the first
condition, $\alpha>0$, is automatically met by the contact damping
model.

--------------------------------------
$I \!=\!\mbox{\scriptsize $\matwo{1}{0}{0}{1}$}$ and $J
\!=\!\mbox{\scriptsize $\matwo{0}{-1}{1}{\,0}$}$ and can be found
in~\cite{qiao_compliance}. Lemma~\ref{lemm.curvature} (in
Appendix~\ref{app.dyn})

In fully active grasping systems natural compliance effects are
dominated by the compliance introduced at the grasping mechanism's
joints~\cite{bicchi_survey,cutkosky&kao}. However, many grasping
and most fixturing systems are under-actuated, and in such systems
the combination of natural compliance and friction effects plays
an important role. For example, in whole arm manipulation an
object is manipulated by one or more articulated mechanisms which
are allowed to establish mid-link contacts with the manipulated
object~\cite{harada&kaneko,omata&nagata}. Due to the passive
nature of the mid-link contacts, the stability of whole-arm grasps
requires a consideration of the natural friction-compliance
contact laws~\cite{xiong&wang05}. Another example is multi-finger
grasping with soft fingertips, where the fingertips' natural
compliance plays a dominant role in the overall grasp
compliance~\cite{montana_icra88,kao99}. A third example is
light-duty fixturing applications for packaging and assembly. In
these applications simple fixturing elements are preloaded against
a workpiece and then passively respond to external loads based on
natural friction-compliance effects~\cite{campbell,barash89,
hockenberger&demeter:1995,hoffman,wang01}.

The paper formulates a rule for preloading frictional grasps and
fixtures which guarantees stable response to external disturbances
at the individual contacts. Then it provides a criterion for
selecting the contact locations which guarantees overall grasp and
fixture stability.

We assume that the grasp arrangement lies on a horizontal plane
such that the finger forces act horizontally on $\Bs$. Under this
assumption $\Bs$'s motions take place within a horizontal plane.
The c-space of $\Bs$ .

obtained by first loading along a linear profile toward
$(\Dg_i^t(q_0),\Dg_i^n(q_0))$, then loading along \noindent the
symmetrized $2\times 2$ matrix \mbox{\scriptsize
$\matwo{\frac{4}{3 (2-\nu)}}{-\frac{2 \mbox{\small $c_i$}}{3
(2-\nu)}}{0}{\frac{1}{1-\nu}}$} at the core of $\tilde{K}_i(q_0)$
is positive definite, and consequently
$\big(\tilde{K}_i(q_0)\big)_s$ is positive semi-definite.
\label{prop.K.p.d} \

(f_i^n I \!+\! f_i^t J)

Let $\Bs$ be grasped at an equilibrium configuration
$q_0=(d_0,\tg_0)$ by stationary bodies $\As_1,\dots,\As_k$. Let
$\DG p_1=q \!-\! q_0$ and $\DG p_2=\qd \!-\! 0$. Then the
linearized dynamics of $\Bs$ at $(q_0,0)$ is given by
 \beq{lin_dyn}
 \frac{d}{dt} \mbox{\small $\left(\!\!\!\! \begin{array}{c } \Delta p_1
\\ \Delta p_2 \end{array} \!\!\!\! \right)
\! =\! \left[\!\! \begin{array}{c c }
     O_{3\times 3} &    I_{3\times 3}   \\
     -M^{-1}(q_0) K_p(q_0,0) & -M^{-1}(q_0) K_d(q_0,0)
\end{array} \!\!\!\right]
\!\! \left(\!\!\!\!  \begin{array}{c } \Delta p_1 \\
\Delta p_2
\end{array} \!\!\!\! \right)\!,
 $}
 \eeq
where $K_p(q_0,0)$ and $K_d(q_0,0)$ are the grasp's stiffness and
damping matrices. The grasp stiffness matrix is the $3\stimes 3$
matrix:
\[
 K_p(q_0,0) = - \sum_{i=1}^k \left\{ \tilde{K}_i(q_0) -
 \mbox{\small $\matwo{O_{2\times 2}}{0}{0^T}{\vrho_i(\tg_0)\cdot f_i}$}
 + G_i^T(q_0) (f_i^n I \!+\! f_i^t J) Dn_i(q_0)  \right\}\!,
\]
\noindent where $\tilde{K}_i(q_0)$ is the c-space stiffness matrix
induced by the $i^{th}$ contact,  $\vrho_i(\tg_0) = R(\tg_0) r_i$,
$f_i$ is the $i^{th}$ contact force with components
$(f_i^t,f_i^n)$, and $n_i$ is the $i^{th}$ contact normal {\rm
($i=1\ldots k$)}. The grasp damping matrix is the $3\stimes 3$
matrix:
\[
K_d(q_0,0) = \sum_{i=1}^k  \mbox{\small $\Dg_i^n(q_0)$} G_i^T(q_0)
R_i(q_0) \matwo{\frac{\partial \varphi_i^t}{\partial
\dot{\delta}_i^t}\mbox{\small $(0)$}}{0}{0}{\frac{\partial
\varphi_i^n}{\partial \dot{\delta}_i^n}\mbox{\small $(0)$}}
R_i^T(q_0) G_i(q_0),
\]
\noindent where $\varphi_i^t$ and $\varphi_i^n$ are the $i^{th}$
tangential and normal damping functions {\rm ($i=1\ldots k$)}.

The stability of asymmetric linear systems has been recently
considered by the first author~\cite{ShapiroAsymmetry2005}.

\begin{thm}[Grasp stability] \label{thm.grasp_stability}
Let a quasi-rigid body $\Bs$ be grasped at an equilibrium
configuration $q_0$ by quasi-rigid bodies $\As_1,\ldots,\As_k$.
Define the scalars $\Ag$, $\Bg$, and $\Gg$ as
\[
\begin{array}{l}
1. \hspace{.5em} \alpha=\lambda_{min}\left(M^{-1/2} (\sum_{i=1}^k
J_i^T R_i K_i R_i^T J_i-D^2 X_{r_i}(q)R_i
\tilde{F}_i-J_i^TDR_i(q)\tilde{F}_i)_s M^{-1/2}\right)\\[2pt]
2. \hspace{.5em} \beta=\lambda_{min}\left(M^{-1/2}(q_0) K_d(q_0,0)
M^{-1/2}(q_0)\right)
> 0 \mbox{where $K_d(q_0,0) = \sum_{i=1}^k  \mbox{\small $\Dg_i^n(q_0)$} G_i^T(q_0)
R_i(q_0) \matwo{\frac{\partial \varphi_i^t}{\partial
\dot{\delta}_i^t}\mbox{\small $(0)$}}{0}{0}{\frac{\partial
\varphi_i^n}{\partial \dot{\delta}_i^n}\mbox{\small $(0)$}}
R_i^T(q_0) G_i(q_0)$}\\[2pt]
3. \hspace{.5em} \Gg=\norm{ M^{-1/2} (\sum_{i=1}^k J_i^T R_i K_i
R_i^T J_i-D^2 X_{r_i}(q)R_i
\tilde{F}_i-J_i^TDR_i(q)\tilde{F}_i)_{as} M^{-1/2}}.
\end{array}
\]
Then if $\Ag > 0$ and $\absvl{\Gg} < \sqrt{\alpha} \beta$, the
zero-velocity state $(q_0,0)$ of the nonlinear grasp
system~\eq{nonlinear_sys} is {\bf locally asymptotically stable.}
\end{thm}

, $\frac{d}{dt}\Dg x = DF(x_0) \Dg x$

Global asymptotically stability of the linearized system about
$(q_0,0)$ implies local asymptotic stability of the original
nonlinear grasp system.

of a~hyperbolic equilibrium\footnote{The system $\frac{d}{dt} x=
F(x)$ such that $f(x_0)=\vec{0}$ has a {\em hyperbolic}
equilibrium when the Jacobian $D F(x_0)$ has eigenvalues with
non-zero real part~\cite{hirsch&smale}.}

In our case the conditions on $\Ag$, $\Bg$, and $\Gg$ ensure
global asymptotic stability of the linearized dynamics at
$(q_0,0)$. This dictates that $(q_0,0)$ is hyperbolic, and
consequently the nonlinear grasp dynamics is locally
asymptotically stable at $(q_0,0)$.

Recall from equation (\ref{eq.lin}) the linearized system about
the equilibrium as follows: According to the Lemma the conditions
for such system to be asymptotically stable are as follows. In
order to satisfy the \textbf{second condition} we need $M^{-1/2}
(K_p)_s M^{-1/2}$ to be positive definite.

Recall from (\ref{eq.eq.K_p_final}) that
\[
(K_p)_s=\left(DG(q)+\sum_{i=1}^k J_i^T R_i K_i R_i^T J_i-D^2
X_{r_i}(q)R_i \tilde{F}_i-J_i^TDR_i(q)\tilde{F}_i \right)_s.
\]
%(\tilde{K})_s=\left(P + \sum_{i=1}^m \left( J_i^T R_i^T K_i R_i
%J_i - D^2x_i(\hat{q}_1)F_i \right) \right)_s.
%\]
$K_i$ can be decomposed as follows,
\[
K_i=(K_i)_s+(K_i)_{as}.
\]
Thus the term $J_i^T R_i^T (K_i)_s R_i J_i$ is symmetric since
\[
\left(J_i^T R_i^T (K_i)_s R_i J_i\right)^T=J_i^T R_i^T
\left((K_i)_s\right)^T R_i J_i= J_i^T R_i^T (K_i)_s R_i J_i,
\]
and $J_i^T R_i^T (K_i)_{as} R_i J_i$ is skew-symmetric since
\[
\left(J_i^T R_i^T (K_i)_{as} R_i J_i\right)^T=J_i^T R_i^T
\left((K_i)_{as} \right)^T R_i J_i= - J_i^T R_i^T (K_i)_{as} R_i
J_i.
\]
The term $D^2 X_{r_i}(q)R_i \tilde{F}_i$ is symmetric because of
the following. $D^2 X_{r_i}(q)R_i \tilde{F}_i$ is the derivative
of $J_i^T R_i F_i$ with respect to $q$, while holding $R_i F_i$
constant. Now we specifically write
\[
J_i^T R_i F_i=\left[ \begin{array}{c c}
\parder{x_1}{q_1} & \parder{x_2}{q_1} \\
\vdots & \vdots \\
\parder{x_1}{q_n} & \parder{x_2}{q_n} \\
\end{array}
\right] \vctwo{f_1}{f_2} = \vcthree{\sum
\parder{x_i}{q_1}f_i}{\vdots}{\sum \parder{x_i}{q_n}f_i},
\]
where for simplicity we denote here $X_i=(x_1,x_2)$ and $R_i
F_i=(f_1,f_2)$. Next we take this term's derivative as follows,
\[
\frac{\partial}{\partial q}\vcthree{\sum
\parder{x_i}{q_1}f_i}{\vdots}{\sum \parder{x_i}{q_n}f_i} = \left[ \begin{array}{c c c}
\sum \frac{\partial^2 x_i}{\partial q_1^2}f_i & \cdots & \sum
\frac{\partial^2 x_i}{\partial q_1 \partial q_n}f_i \\
\vdots &  & \vdots \\
\sum \frac{\partial^2 x_i}{\partial q_n \partial q_1}f_i & \cdots
& \sum
\frac{\partial^2 x_i}{\partial q_n^2}f_i \\
\end{array}
\right]
\]
It is easy to see that this matrix is symmetric.

\[
-\sum_{i=1}^k \ve^T (\tilde{K}_i(q_0))_s \ve + \sum_{i=1}^k
\vrho_i(\tg_0)\cdot
 f_i > 0,
 \]
where $\ve \!=\! (0,0,1)$, $f_i$ is the $i^{th}$ finger force, and
$\vrho_i(\tg_0) \!=\! R(\tg_0) r_i$ is the vector from $\Bs$'s
origin to $x_i$ ($i=1\ldots k$).

The positive definiteness of $M^{-1/2} (K_p)_s M^{-1/2}$ follows
from the positive definiteness of $(K_p)_s$. Using
eq.~\eq{k_simple},

Since $q_0$ is an equilibrium configuration, $\, f_{ext}=
-(\sum_{i=1}^k f_i(q) + f_g)$. Substituting for $f_{ext}$ in the
lower-right term of~\eq{pdi} gives the result.

First, their contact model requires that the contacts be preloaded
only along the contact normals. The Hertz-Walton model allows any
linear loading of the contacts. Second, they judge the stability
of a candidate grasp by constructing a potential energy function
and computing its Hessian matrix. While a potential-function
approach may be justified under normal loading profiles, this
paper establishes that under general linear loading profiles the
resulting grasp is not governed by any potential energy function.
Third, the current paper emphasizes the role of the contact
loading profile in ensuring grasp stability, a~topic which has not
yet been discussed in the robotics~literature.

 (i.e. the
eigenvalues of the $4\! \times\! 4$ matrix lie in the left half of
the complex plane)

$\vrho_i(\tg_0)$ is the vector from $\Bs$'s origin to the $i^{th}$
contact point,

(Example~1). As discussed in Section~\ref{sec.stiffness}, the
individual matrices $(\tilde{K}_i)_s$ are negative definite for
linear loading profiles and reasonable friction coefficients.
Hence the only potentially destabilizing term is the sum
$\sum_{i=1}^k \vrho_i(\tg_0)\cdot f_i$, which corresponds to the
effect of the preload forces.

 Let a quasi-rigid $\Bs$ be held at an equilibrium grasp
configuration $q_0 \!=\! (d_0,\tg_0)$ by quasi-rigid bodies
$\As_1,\dots,\As_k$. Let $\DG p_1=q \!-\! q_0$ and $\DG p_2=\qd
\!-\! 0$. The linearized dynamics of $\Bs$ at $(q_0,0)$ is given
by
 \beq{lin_dyn_app}
 \frac{d}{dt} \mbox{\small $\left(\!\!\!\! \begin{array}{c } \Delta p_1
\\ \Delta p_2 \end{array} \!\!\!\! \right)
\! =\! \left[\!\! \begin{array}{c c }
     O_{3\times 3} &    I_{3\times 3}   \\
     -M^{-1}(q_0) K_p(q_0,0) & -M^{-1}(q_0) K_d(q_0,0)
\end{array} \!\!\!\right]
\!\! \left(\!\!\!\!  \begin{array}{c } \Delta p_1 \\
\Delta p_2
\end{array} \!\!\!\! \right)\!,
 $}
 \eeq
where $K_p$ and $K_d$ are the following grasp's stiffness and
damping matrices. The grasp's stiffness matrix is the asymmetric
$3\times 3$ matrix:
\[
 K_p(q_0,0) = - \sum_{i=1}^k \left\{ \tilde{K}_i(q_0) -
 \mbox{\small $\matwo{O_{2\times 2}}{0}{0^T}{\vrho_i(\tg_0)\cdot f_i}$}
 + G_i^T(q_0) \matwo{f_i^n}{-f_i^t}{f_i^t}{f_i^n} Dn_i(q_0)  \right\}\!,
\]
\noindent where $\tilde{K}_i(q_0)$ is the c-space stiffness matrix
induced by the $i^{th}$ contact,  $\vrho_i(\tg_0) = R(\tg_0) r_i$,
$f_i$ is the $i^{th}$ contact force with components
$(f_i^t,f_i^n)$, and $n_i$ is the $i^{th}$ contact normal {\rm
($i=1\ldots k$)}. The grasp's damping matrix is the symmetric
$3\stimes 3$ matrix:
\[
K_d(q_0,0) = \sum_{i=1}^k  \mbox{\small $\Dg_i^n(q_0)$} G_i^T(q_0)
R_i(q_0) \matwo{\frac{\partial \varphi_i^t}{\partial
\dot{\delta}_i^t}\mbox{\small $(0)$}}{0}{0}{\frac{\partial
\varphi_i^n}{\partial \dot{\delta}_i^n}\mbox{\small $(0)$}}
R_i^T(q_0) G_i(q_0),
\]
\noindent where $\varphi_i^t$ and $\varphi_i^n$ are the tangential
and normal damping functions at the $i^{th}$ contact {\rm
($i=1\ldots k$)}.}

\]
where  are the preload forces at the contacts $x_1,\ldots,x_k$.

\vspace{.125in}\noindent  \textbf{Example:} Consider the
two-finger grasp of an ellipse depicted in
Figure~\ref{passive_ex.fig}, which takes place in a horizontal
plane without gravity. When the ellipse is grasped by the
compliant fixturing devices depicted in the figure, the contacts
satisfy linear force-displacement laws having stiffness matrices
$K_1=K_2= \mbox{\scriptsize $\matwo{\Kg}{0}{0}{\Kg}$}$ for some
$\Kg>0$. The grasp's stability depends on the elastic energy
stored in the compressed contacts. Grasp stability requires that
the equilibrium configuration be a strict local minimum of the
grasp's total potential energy function. Beyond a certain preload
level this grasp ceases to be a local minimum of its potential
energy function, and the grasp is no longer stable. This
phenomenon is known as \qt{coin snapping} in the grasping
literature~\cite{cutkosky&kao,qiao_compliance,montana92}. The
preload forces should therefore be sufficiently small so that they
would not harm the positive definiteness of $(K_p)_s$ and
consequently the positive definiteness of $M^{-1/2} (K_p)_s
M^{-1/2}$.


Figure~\ref{penetrations_constraints.fig} depicts

 the restriction
imposed on the linear loading profile by~\eq{slope}. Since $0\leq
\nu \leq 0.5$, the slope of the linear loading path must satisfy
$\absvl{c_i} < 6.0$ for $(\tilde{K}_i)_s$ to be negative
semi-definite. The corresponding slope angle, denoted $\beta$ in
Figure~\ref{penetrations_constraints.fig}, must satisfy
$\absvl{\beta} < 80.5^\circ$. However, $c_i$ must also satisfy the
friction cone constraint specified in~\eq{walton},  $\absvl{c_i} <
\mu (2-\nu)/2(1-\nu)$. For typical values friction coefficients in
the range $\mu \leq 1$, the linear loading path's slope must
satisfy the inequality $\absvl{c_i} < 1.5$. The corresponding
slope angle, denoted $\alpha$ in
Figure~\ref{penetrations_constraints.fig}, must satisfy
$\absvl{\alpha} < 56.3^\circ$. It follows that the negative
semi-definiteness of $(\tilde{K}_i)_s$ is typically {\em less
restrictive} than the friction cone constraint. Since the linear
loading path must always satisfy the friction cone constrain, the
matrix $(\tilde{K}_i)_s$ is automatically negative semi-definite
in typical linearly loaded grasps.

\noindent Since $t_i= J n_i$, the matrix $R_i= [ t_i \, n_i ]$ can
be written as $R_i= [ J n_i \, n_i ]$, where  $J
\!=\!\mbox{\scriptsize $\matwo{0}{-1}{1}{\,0}$}$. Hence the first
summand of~\eq{K_p} can be written as $DR(q_0)= [ J Dn_i(q_0) \,
Dn_i(q_0) ]$, and $\big(D R_i(q_0) \big) \mbox{\scriptsize
$\vctwo{f_i^t}{f_i^n}$} = ( f_i^t J + f_i^n I) Dn_i(q_0)$, where
$I$ is a $2\times 2$ identity matrix. Let us now compute the
second summand in~\eq{K_p}. According to the Hertz-Walton model,
the force components $(f_i^t,f_i^n)$ are given by
 \beq{vf}
 \vctwo{f_i^t (\mbox{\small
$\vect{\Dg}_i,\dot{\vect{\Dg}}_i$} )}{f_i^n (\mbox{\small
$\vect{\Dg}_i,\dot{\vect{\Dg}}_i$} )} =
\vctwo{-h_i(\Dg_i^t,\Dg_i^n)}{g_i(\Dg_i^n)} + \vctwo{-\Dg_i^n
\!\cdot\! \Pg_i^t(\mbox{\small $\dot{\Dg}_i^t$})}{\Dg_i^n
\!\cdot\! \Pg_i^n(\mbox{\small $\dot{\Dg}_i^n)$}}\!.
 \eeq
\noindent The derivative of $(f_i^t,f_i^n)$ with respect to
$\vect{\Dg}_i$ at $(q_0,0)$ is given by
\[
 \att{\vctwo{\frac{\partial}{\partial \vect{\Dg}_i}
f_i^t(\mbox{\small $\vect{\Dg}_i,\dot{\vect{\Dg}}_i$})}{
\frac{\partial}{\partial \vect{\Dg}_i} f_i^n(\mbox{\small
$\vect{\Dg}_i,\dot{\vect{\Dg}}_i$})}}{\!\!\!\mbox{\tiny
$\begin{array}{l} p_1 \!=\! q_0 \\ p_2 \!=\! 0
 \end{array}$}}
 \!\!\!= K_i(q_0) +
\att{\matwo{-\Pg_i^t(\mbox{\small
$\dot{\Dg}_i^t$})}{0}{\Pg_i^n(\mbox{\small $\dot{\Dg}_i^n)$}}{0}}{
\!\!\!\mbox{\tiny $\begin{array}{l} p_1 \!=\! q_0 \\ p_2 \!=\! 0
 \end{array}$}}
\!\!\!= K_i(q_0),
\]
where we used the fact that $\dot{\Dg}_i^t \!=\! \dot{\Dg}_i^n
\!=\! 0$ at $p_2=0$, and consequently  $\Pg_i^t(0)=\Pg_i^n(0)=0$.
The Jacobian $D \vect{\Dg}_i(p_1)$ appearing in the second summand
of~\eq{K_p} is given by $D\vect{\Dg}_i= \mbox{\scriptsize $\left[
\begin{array}{c }
  t_i^T  \\ -n_i^T \end{array} \right]$} G_i$ (see proof of
Lemma~\ref{lemm.stiffness_matrix}). As for the third summand
in~\eq{K_p}, $\dot{\vect{\delta}}_i= \mbox{\scriptsize $\left[
\begin{array}{c}
  t_i^T  \\ -n_i^T \end{array} \right]$} G_i \qd$, and consequently
$\frac{\partial}{\partial p_1 }\dot{\vect{\Dg}}_i(p_1,p_2)=0$ at
$p_2=0$. Substituting these terms in~\eq{K_p}, then summing the
expressions $D^2 X_{r_i}^T(p_1) f_i(p_1,p_2) + G_i^T
\frac{\partial}{\partial p_1} f_i(p_1,p_2)$ over $i=1\ldots k$
gives
\[ \barr
K_p(q_0,0) =& \left. -\frac{\partial}{\partial p_1} \right|_{\tiny
\!\!\!\! \begin{array}{l} p_1 \!=\! q_0 \\ p_2 \!=\! 0
 \end{array}}
\sum_{i=1}^k G_i^T(p_1) f_i(p_1,p_2) \\[2pt]
& =  -\sum_{i=1}^k \left\{ G_i^T(q_0) R_i(q_0) K_i(q_0)
\mbox{\scriptsize $\left[
\begin{array}{c } t_i^T  \\
-n_i^T \end{array} \right]$} G_i(q_0) \right.\\[2pt]
& \left. - \mbox{\small $\matwo{O_{2\times
2}}{0}{0^T}{\vrho_i(\tg_0)\cdot f_i}$}
 + G_i^T(q_0) (f_i^n I \!+\! f_i^t J) Dn_i(q_0)
\right\}\!, \earr
\]
where we substituted $D^2 X_{r_i}^T(q_0) f_i= \mbox{\scriptsize
$\matwo{O}{0}{0^T}{-\vrho_i\cdot f_i}$}$ in accordance with
Lemma~\ref{lemm.DX_ri}. Based on the proof of
Lemma~\ref{lemm.stiffness_matrix}, $\tilde{K}_i= G_i^T(q_0)
R_i(q_0) K_i(q_0) \mbox{\scriptsize $\left[
\begin{array}{c }
  t_i^T  \\ -n_i^T \end{array} \right]$} G_i(q_0)$.
Substituting for this term gives the formula
\[
K_p(q_0,0) =  -\sum_{i=1}^k \left\{ \tilde{K}_i(q_0)
 - \mbox{\small $\matwo{O_{2\times 2}}{0}{0^T}{\vrho_i(\tg_0)\cdot f_i}$}
 + G_i^T(q_0) (f_i^n I \!+\! f_i^t J) Dn_i(q_0)
 \right\}\!.
\]
\noindent Finally consider the matrix $K_d(q_0,0)$. We have to
compute the derivative:
\[
\att{\frac{\partial }{\partial p_2}}{ \!\!\!\mbox{\tiny
$\begin{array}{l} p_1 \!=\! q_0 \\ p_2 \!=\! 0
 \end{array}$}} G_i^T(p_1) f_i(p_1,p_2)=
G_i^T(p_1) \att{\frac{\partial }{\partial p_2}}{ \!\!\!\mbox{\tiny
$\begin{array}{l} p_1 \!=\! q_0 \\ p_2 \!=\! 0
 \end{array}$}} \! f_i(p_1,p_2).
 \]
Based on \eq{vf},
\[
 \frac{\partial }{\partial p_2} f_i(p_1,p_2) = R_i(p_1)
 \matwo{-\Dg_i^n(p_1) \!\cdot\! \frac{\partial
}{\partial \dot{\Dg}_i^t}\varphi_i^t(\mbox{\small
$\dot{\Dg}_i^t$})}{0}{0}{\Dg_i^n(p_1) \!\cdot\!
\frac{\partial}{\partial \dot{\Dg}_i^n}\varphi_i^n(\mbox{\small
$\dot{\Dg}_i^n$})} \frac{\partial }{\partial p_2}
\dot{\vect{\Dg}}_i(p_1,p_2).
\]
Since  $\dot{\vect{\delta}}_i= \mbox{\scriptsize $\left[
\begin{array}{c}
  t_i^T  \\ -n_i^T \end{array} \right]$} G_i p_2$, we have that
$\frac{\partial }{\partial p_2} \dot{\vect{\delta}}_i(p_1,p_2) =
\mbox{\scriptsize $\left[
\begin{array}{c}
  t_i^T  \\ -n_i^T \end{array} \right]$} G_i$.
Hence
\[ \barr
\att{\frac{\partial }{\partial p_2}}{ \!\!\!\mbox{\tiny
$\begin{array}{l} p_1 \!=\! q_0 \\ p_2 \!=\! 0
 \end{array}$}} f_i(p_1,p_2) &= \Dg_i^n(q_0)  R_i(q_0) \matwo{-\frac{\partial
\varphi_i^t}{\partial \dot{\delta}_i^t}(0)}{0}{0}{\frac{\partial
\varphi_i^n}{\partial \dot{\delta}_i^n}(0)} \mbox{\scriptsize
$\left[
\begin{array}{c}
  t_i^T  \\ -n_i^T \end{array} \right]$} G_i(q_0) \\[2pt]
& = - \Dg_i^n(q_0) R_i(q_0) \matwo{ \frac{\partial
\varphi_i^t}{\partial \dot{\delta}_i^t}(0)}{0}{0}{\frac{\partial
\varphi_i^n}{\partial \dot{\delta}_i^n}(0)} R_i^T(q_0) G_i(q_0),
\earr
\]
where we substituted $R_i^T = \mbox{\scriptsize $\left[
\begin{array}{c}
  t_i^T  \\ n_i^T \end{array} \right]$}$. Summing over the $k$ contacts,
\[ \barr
K_d(q_0,0) &= \left. -\frac{\partial}{\partial p_2}\right|_{\tiny
\!\!\!\! \begin{array}{l} p_1 \!=\! q_0 \\ p_2 \!=\! 0
\end{array}} \sum_{i=1}^k G_i^T(p_1) f_i(p_1,p_2) \\
&= \sum_{i=1}^k  \Dg_i^n(q_0) G_i^T(q_0) R_i(q_0)
\matwo{\frac{\partial \varphi_i^t}{\partial
\dot{\Dg}_i^t}(0)}{0}{0}{\frac{\partial \varphi_i^n}{\partial
\dot{\Dg}_i^n}(0)} R_i^T(q_0) G_i(q_0). \earr
\]
 \epf

\left[ O \!, G_i^T(q_0)  R_i \vctwo{f_i^n}{f_i^t} \, \right]
 \right\}\!.

\[
\sum_{i=1}^k G_i^T(q_0) \matwo{f_i^n}{f_i^t}{-f_i^t}{f_i^n}
Dn_i(q_0) = \sum_{i=1}^k \matwo{O_{2\times
2}}{0}{0^T}{\sum_{i=1}^k x_i\cdot f_i}\!.
\]

\[
\sum_{i=1}^k G_i^T(q_0) \matwo{f_i^n}{f_i^t}{-f_i^t}{f_i^n}
Dn_i(q_0) = \sum_{i=1}^k \matwo{O_{2\times
2}}{0}{0^T}{\sum_{i=1}^k x_i\cdot f_i}\!.
\]

where $t_i = J^T n_i$ is $\Bs$'s unit tangent at $x_i$

 for $J= \mbox{\scriptsize
$\matwo{0}{-1}{1}{0}$ }$.

Since $R_i=[t_i \, n_i]$ and $R_i J = J R_i$, one obtains
\[
\barr & \sum_{i=1}^k G_i^T(q_0)
\matwo{f_i^n}{f_i^t}{-f_i^t}{f_i^n} Dn_i(q_0) = \left[ O_{3\times
2},
\sum_{i=1}^k  G_i^T(q_0)  R_i \vctwo{-f_i^n}{f_i^t} \, \right]\\
& = \left[ O_{3\times 2}, \sum_{i=1}^k G_i^T(q_0)  R_i J
\vctwo{f_i^t}{f_i^n} \, \right] = \left[ O_{3\times 2},
\sum_{i=1}^k G_i^T(q_0)  J f_i \, \right] \!, \earr
\]
where $f_i = R_i \mbox{\scriptsize $\vctwo{f_i^t}{f_i^n}$}$ is the
$i^{th}$ contact force.

+
 \mbox{\small $\matwo{O_{2\times 2}}{0}{0^T}{2\sum_{i=1}^k x_i\cdot
 f_i}$} \right\}.

 \begin{lemm}[\bf \boldmath{$\Ag \!>\! 0$} test] \label{lemm.posdef}
Let $\Bs$ be held in a compliant equilibrium grasp with c-space
contact stiffness matrices $\tilde{K}_i$ for $i \!=\! 1\ldots k$.
Write the symmetric $3\times 3$  matrix $-\! \sum_{i=1}^k
\big(\tilde{K}_i(q_0)\big)_s$ as
\[
-\! \sum_{i=1}^k \big(\tilde{K}_i(q_0)\big)_s =
\matwo{P_{11}}{P_{12}}{P_{12}^T}{P_{22}}\!.
\]
Then $\Ag \!>\! 0$ when $-\!\sum_{i=1}^k
\big(\tilde{K}_i(q_0)\big)_s$ is positive definite, and the
preload forces $f_1,\ldots,f_k$ satisfy the inequality:
$\sum_{i=1}^k x_i \cdot f_i < P_{22} - P_{12}^T P_{11}^{-1}
P_{12}$.
 \elemm

\bpf First express the matrix $(K_p)_s$ in terms of $P_{11}$,
$P_{12}$, and $P_{22}$:
\[
(K_p(q_0,0))_s = \matwo{P_{11}}{P_{12}}{P_{12}^T}{P_{22}} +
\matwo{O}{0}{0^T}{\sum_{i=1}^k x_i \!\cdot\! f_i(q)} \!.
\]
Consider now the transformation of the first summand:\footnote{The
transformation is induced by a~translation of $\Bs$'s reference
frame to the grasp's center of compliance.}
\[
\matwo{I}{-P_{11}^{-1}P_{12}}{0}{1}^T
\matwo{P_{11}}{P_{12}}{P_{12}^T}{P_{22}}
\matwo{I}{-P_{11}^{-1}P_{12}}{0}{1} =
\matwo{P_{11}}{0}{0^T}{P_{22} - P_{12}^T P_{11}^{-1}P_{12}}\!.
\]
Since the second summand of $(K_p)_s$ is invariant under this
transformation, $(K_p)_s$ is related by the transformation to the
block-diagonal matrix:
 \beq{pdi}
(K_p(q_0,0))_s  \sim \matwo{P_{11}}{0}{0^T}{P_{22} - P_{12}^T
P_{11}^{-1}P_{12} + \sum_{i=1}^k x_i \!\cdot\! f_i }\!.
 \eeq
The eigenvalues of $(K_p)_s$ have the same sign as the eigenvalues
of the block diagonal matrix in~\eq{pdi}. Hence the condition
$\Lg_{min} \prl{(K_p)_s}
>0$ is equivalent to the positiveness definiteness of the block
diagonal matrix in~\eq{pdi}. Since the matrix $-\!\sum_{i=1}^k
\big(\tilde{K}_i\big)_s$  is positive definite, its upper left
$2\!\times\! 2$ block, $P_{11}$, is positive definite. Hence
$(K_p)_s$  is positive definite when $P_{22} \!-\! P_{12}^T
P_{11}^{-1}P_{12} + \sum_{i=1}^k x_i \!\cdot\! f_i>0$.
 \epf


\vspace{.125in}\noindent A proof of the lemma appears in the
appendix. The lemma is based on the fact that planar grasps
possess a special point, called the {\em center of compliance,}
such that when $\Bs$'s origin is selected at this point the
grasp's stiffness matrix becomes block diagonal~\cite{nguyen}. The
term $P_{22} - P_{12}^T P_{11}^{-1} P_{12} \!-\! \sum_{i=1}^k x_i
\cdot f_i$ is the grasp's rotational stiffness about its center of
compliance. The preload forces can destabilize the grasp when
$\sum_{i=1}^k x_i \!\cdot\! f_i$ is sufficiently negative, as
illustrated in the following example.

\vspace{.125in}\noindent  \textbf{Example:} Consider the
two-finger grasp of a rectangular object depicted in
Figure~\ref{passive_ex.fig}, which takes place in a horizontal
plane without gravity. The fingers are preloaded along the contact
normals, which correspond to zero-slope loading profiles.
Substituting $c_1=c_2=0$ for in~\eq{inspect}, the contact
stiffness matrices are given by
\[
 K_i(q_0) = \mbox{\small $4 G
$} \sqrt{\Dg_i^n(q_0)} \matwo{-\frac{4}{3
(2-\nu)}}{0}{0}{\frac{1}{1-\nu}}\!.
\]
The corresponding c-space contact stiffness matrices are given by
\[
\tilde{K}_i(q_0) = -\mbox{\small $4 G $} \sqrt{\Dg_i^n(q_0)}
G_i^T(q_0) \matwo{\frac{4}{3 (2-\nu)}}{0}{0}{\frac{1}{1-\nu}}
G_i(q_0).
 \]
The grasp's stability depends on the elastic energy stored in the
compressed contacts. Grasp stability requires that the equilibrium
configuration be a strict local minimum of the grasp's total
potential energy function. Beyond a certain preload level this
grasp ceases to be a local minimum of its potential energy
function, and the grasp is no longer stable. This phenomenon is
known as \qt{coin snapping} in the grasping
literature~\cite{cutkosky&kao,qiao_compliance,montana92}. The
preload forces should therefore be sufficiently small so that they
would not harm the positive definiteness of $(K_p)_s$ and
consequently the positive definiteness of $M^{-1/2} (K_p)_s
M^{-1/2}$.

%%%%-----------------------------------------------------------------------%%%
\begin{figure}
\centering{\includegraphics[width=0.52\textwidth]{passive_preload.eps}}
\vspace{-.125in} \caption{A frictional two-finger grasp which
ceases to be stable beyond a certain preload level.}
\label{passive_ex.fig}
\end{figure}
%%%-----------------------------------------------------------------------%%%

In general, when $Q$ and $P$ are $n\times n$ matrices such that
$Q$ is non-singular and $P$ symmetric, the eigenvalues of $Q^T P
Q$ have the same sign as the eigenvalues of $P$.

where $A_i \!=\! \mbox{\scriptsize $\matwo{\frac{4}{3
(2-\nu)}}{-\frac{2 \mbox{\small $c$}}{3
(2-\nu)}}{0}{\frac{1}{1-\nu}}$}$.


$r_i \!=\! r_{1i}
r_{2i} / ( r_{1i} \!+\! r_{2i})$,



The grasp stiffness
matrix becomes
\[
K_p(q_0,0) = -\! \sum_{i=1}^k \tilde{K}_i(q_0) = \sm{\sqrt{\Dg_c^n}}
\sum_{i=1}^k G_i^T(q_0) R_i(q_0) A_i R_i^T(q_0) G_i(q_0).
 \]

 As
discussed in Section~\ref{sec.stab_analysis}, \eq{3cond}
essentially means that $(K_p)_{as}$ should be sufficiently small
relative to $(K_p)_s$ (which is modulated by $\Kg_d$).

$K_p = -\! \sum_{i=1}^k \tilde{K}_i(q_0)$

\[
\sum_{i=1}^k S^T G_i^T(q_0) G_i(q_0) S = k \matwo{I}{0}{0^T}{\frac{1}{k l_c^2}\sum_{i=1}^k \norm{x_i}^2}
=\!,
\]
where we substituted . Thus

\bpf
First consider the term $\Ag=\Lg_{min}(\bar{M}^{-1/2} (\bar{K}_p)_s \bar{M}^{-1/2})$. The minimal eigenvalue of $M^{-1/2} (\bar{K}_p)_s
M^{-1/2}$ can be written as
\[
\Lg_{min}( \bar{M}^{-1/2} (\bar{K}_p)_s \bar{M}^{-1/2})= \min_{\norm{v}=1} \left\{
v^T M^{-\shlf} (\bar{K}_p)_s M^{-\shlf}  v \right\} =
\min_{u^T \bar{M} u=1} \left\{ u^T (\bar{K}_p)_s   u \right\},
\]
where $u \!=\! \bar{M}^{-\shlf} v$. Substituting for $(\bar{K}_p)_s$ according to~\eq{ssymm}
gives
\[
\Lg_{min}((\bar{K}_p)_s) \geq   \sm{\sqrt{\Dg_c^n}} \Lg_{min}\big(A_s\big)
\cdot \min_{u^T \bar{M} u=1} \left\{ u^T  \big(  \sum_{i=1}^k S^T G_i^T
G_i S \big) u \right\},
\]
where we substituted  $R_i R_i^T \!=\! I$ for $i \!=\! 1\ldots k$. Without loss of generality, let the world frame and $\Bs$'s body frame be located at the
the grasp's centroid. Under this assumption the position of the finger contacts
in $\Bs$'s frame is identical to their position in the
world frame, $x_i \!=\! \vrho_i\sm{(\tg_0)}$ for $i \!=\! 1\ldots
k$, while $x_c =\vec{0}$. The $2\times 3$ Jacobian $G_i =DX_{r_i}$ is given
by $G_i = \brl{ I \, J\!x_i}$, where $I$ is a $2\times 2$ identity
matrix and $J \!=\!\mbox{\scriptsize $\matwor{0}{-1}{1}{\,0}$}$. Hence
\[ \barr
\sum_{i=1}^k S^T G_i^T G_i S &=  \sum_{i=1}^k \matwo{I}{0}{0^T}{1/l_c} \left[ \begin{array}{c}
  I        \\
  x_i^T J^T
\end{array} \right] \brl{ I \, J\! x_i} \matwo{I}{0}{0^T}{1/l_c}\\[6pt]
 & = \matwo{k I}{\frac{1}{l_c}J \sum_{i=1}^k x_i}{\frac{1}{l_c}(\sum_{i=1}^k x_i)^T J^T}{\frac{1}{l_c^2}\sum_{i=1}^k \norm{x_i}^2}\\
 & = k \matwo{I}{\frac{1}{l_c}J x_c}{\frac{1}{l_c}x^T_c J^T}{\frac{1}{k l_c^2}\sum_{i=1}^k \norm{x_i}^2} =  k I_{3\times 3},
  \earr
 \]
where we substituted $x_c \!=\! \frac{1}{k}\sum_{i=1}^k x_i = \vec{0}$ and $l_c^2 \!=\! \frac{1}{k} \sum_{i=1}^k \norm{x_i}^2$.
The matrix  $\sum_{i=1}^k S^T G_i^T G_i S$ is thus essentially the identity matrix, and the lower bound on $\Lg_{min}\big((\bar{K}_p)_s\big)$ is given by
\[
\Lg_{min}\big((\bar{K}_p)_s\big) \geq   k \sm{\sqrt{\Dg_c}} \Lg_{min}(A_s).
\]

An upper bound on the left side of~\eq{3cond_scaled} can be
obtained as follows. The norm of $M^{-1/2} (\bar{K}_p)_{as}
M^{-1/2}$ can be written as
\[
\norm{M^{-\shlf} (\bar{K}_p)_{as} M^{-\shlf}} = \max_{\norm{v}=1} \left\{
v^T M^{-\shlf} (\bar{K}_p)_{as} M^{-\shlf}  v \right\} =
\max_{\norm{u}=1} \left\{ u^T (\bar{K}_p)_{as}   u \right\}\!,
\]
where $u \!=\! M^{-\shlf} v$. Substituting for $(\bar{K}_p)_{as}$ according to~\eq{sas}
gives
\[
\norm{(\bar{K}_p)_{as}} \leq   \sm{\sqrt{\Dg_c}} \norm{A_{as}}
\cdot \max_{\norm{u}=1} \left\{ u^T  \big(\sum_{i=1}^k S^T G_i^T
G_i S \big) u \right\}.
\]
Since $\norm{\sum_{i=1}^k S^T G_i^T
G_i S} \!=\! k$, we obtain that $\norm{(\bar{K}_p)_{as}} \leq k \sm{\sqrt{\Dg_c}} \norm{A_{as}}$. The condition $\Gg \!<\! \sqrt{\alpha} \beta$ is thus equivalent to the condition:
\[
\norm{A_{as}} < \Kg_d \Lg_{min}(A_s),
\]
where we canceled the common factor $k \sm{\sqrt{\Dg_c}}$ from both sides of the inequality.
\epf

Based on Lemma~\ref{lemm.eqv_cond}, the condition $\Gg \!<\! \sqrt{\alpha} \beta$ essentially requires that the symmetric part of the contact stiffness matrix, $A_s$, will dominate its asymmetric part, $A_{as}$. The symmetric and asymmetric parts of $A$ are given by
\[
A_s \!=\! \matwo{\frac{4}{3 (2-\nu)}}{-\frac{
c}{3 (2-\nu)}}{-\frac{
c}{3 (2-\nu)}}{\frac{1}{1-\nu}}
\txt{and}
A_{as} \!=\! \matwo{0}{-\frac{c}{3 (2-\nu)}}{\frac{
c}{3 (2-\nu)}}{0}\!.
\]
The norm of $A_{as}$ is given by
\[
\norm{A_{as}} = \Lg_{max} ( A_{as}^T A_{as})= \frac{c^2}{9(2-\nu)^2}.
\]
The minimal eigenvalue of $A_s$ can be lower bounded by
\[
\Lg_{min}(A_s) \geq \frac{\det(A_s)}{\mbox{\sf tr} (A_s)}.
\]
Substituting for the determinant and trace of $A_s$ gives
\[
\Lg_{min}(A_s) \geq \frac{ \frac{4}{3 (2-\nu)(1-\nu)} - \frac{c^2}{9 (2-\nu)^2}}{
\frac{4}{3 (2-\nu)} + \frac{1}{1-\nu}}.
\]

\vspace{1in} And the third condition for stability is
$\Gg=\|M^{-1/2} (K_p)_{as} M^{-1/2}\| < \sqrt{\alpha} \beta$. To
satisfy the \textbf{third condition} we need $\Gg=\|M^{-1/2}
(K_p)_{as} M^{-1/2}\| < \sqrt{\alpha} \beta$. Specifically we have
\[
(K_p)_{as}=\sum_{i=1}^m J_i^T R_i^T (K_i)_{as} R_i J_i,
\]
and recall that
\[
K_i=\matwo{\frac{4 G \sqrt{R \delta_i^n}}{1-\nu}}{0}{\frac{8 G R
\delta_i^t}{3 \sqrt{R \delta_i^n} (2-\nu)}}{\frac{16 G \sqrt{R
\delta_i^n}}{3 (2-\nu)}}.
\]
Substituting Walton loading path
$c_i=\frac{\delta_i^t}{\delta_i^n}$ yields
\[
K_i=G \sqrt{R \delta_i^n} \matwo{\frac{4} {1-\nu}}{0}{\frac{8
c_i}{3 (2-\nu)}}{\frac{16}{3 (2-\nu)}}.
\]
Which its skew-symmetric part is
\[
(K_i)_{as}=G \sqrt{R \delta_i^n}\matwo{0}{-\frac{8 c_i}{3
(2-\nu)}}{\frac{8 c_i}{3 (2-\nu)}}{0}.
\]
Note that the maximum eigenvalue of $(K_i)_{as}$ depends on the
value of $|c_i|$. Small $|c_i|$ correspond to penetrations which
are more in the normal direction and consequently forces that are
more normal than tangential. Moreover, small $|c_i|$ introduces
small asymmetric part of $K_i$ and help satisfy the third
condition of the theorem. Thus the third condition introduce an
upper bound to the sum of the $|c_i|$s. However this is not the
only condition on the $|c_i|$s. From proposition \ref{prop.K.p.d}
we have the following condition for $(K_i)_s$ to be positive
definite:
\[
|c_i|<\sqrt{\frac{12 (2-\nu)}{(1-\nu)}}.
\]
We also have an upper bound on $|c_i|$ which guarantees that the
contact force is within its friction cone as follows:
\[
c_i^f=\frac{F_i^t}{F_i^n}=c_i \frac{2 (1-\nu)}{(2-\nu)}<\mu
\]
Finally, the third upper bound on $|c_i|$ is from the third
condition of Corollary~\ref{cor.grasp_stability}
\[
\|M^{-1/2} (K_p)_{as} M^{-1/2}\| < \sqrt{\alpha} \beta
\]
as discussed before. The selection of $|c_i|$s value is such that
it satisfies all these three upper bounds. This together with the
other two conditions, are sufficient for the equilibrium of the
grasping arrangement to be {\em locally asymptotically stable}.
 \epf



\max_{\norm{v}=1} \left\{
v^T \bar{M}^{-\shlf} (\bar{K}_p)_{as} \bar{M}^{-1} (\bar{K}_p)_{as} \bar{M}^{-\shlf}  v \right\}\!.

where $\Lg_{max}(\bar{M}^{-1})= \max_{\norm{v}=1} \{ v^T \bar{M}^{-1} v \}$.

\[
\Lg_{min}(A_s) \geq \frac{ \frac{4}{3 (1-\nu)(2-\nu)} - \frac{c^2}{9 (2-\nu)^2}}{
\frac{4}{3 (2-\nu)} + \frac{1}{1-\nu}}.
\]

\[
\big( \Kg + \frac{4}{3 (2-\nu)} + \frac{1}{1-\nu} \big) \frac{c^2}{9(2-\nu)^2} < \Kg \frac{4}{3 (1-\nu)(2-\nu)}.
\]

\vspace{1in} And the third condition for stability is
$\Gg=\|M^{-1/2} (K_p)_{as} M^{-1/2}\| < \sqrt{\alpha} \beta$. To
satisfy the \textbf{third condition} we need $\Gg=\|M^{-1/2}
(K_p)_{as} M^{-1/2}\| < \sqrt{\alpha} \beta$. Specifically we have
\[
(K_p)_{as}=\sum_{i=1}^m J_i^T R_i^T (K_i)_{as} R_i J_i,
\]
and recall that
\[
K_i=\matwo{\frac{4 G \sqrt{R \delta_i^n}}{1-\nu}}{0}{\frac{8 G R
\delta_i^t}{3 \sqrt{R \delta_i^n} (2-\nu)}}{\frac{16 G \sqrt{R
\delta_i^n}}{3 (2-\nu)}}.
\]
Substituting Walton loading path
$c_i=\frac{\delta_i^t}{\delta_i^n}$ yields
\[
K_i=G \sqrt{R \delta_i^n} \matwo{\frac{4} {1-\nu}}{0}{\frac{8
c_i}{3 (2-\nu)}}{\frac{16}{3 (2-\nu)}}.
\]
Which its skew-symmetric part is
\[
(K_i)_{as}=G \sqrt{R \delta_i^n}\matwo{0}{-\frac{8 c_i}{3
(2-\nu)}}{\frac{8 c_i}{3 (2-\nu)}}{0}.
\]
Note that the maximum eigenvalue of $(K_i)_{as}$ depends on the
value of $|c_i|$. Small $|c_i|$ correspond to penetrations which
are more in the normal direction and consequently forces that are
more normal than tangential. Moreover, small $|c_i|$ introduces
small asymmetric part of $K_i$ and help satisfy the third
condition of the theorem. Thus the third condition introduce an
upper bound to the sum of the $|c_i|$s. However this is not the
only condition on the $|c_i|$s. From proposition \ref{prop.K.p.d}
we have the following condition for $(K_i)_s$ to be positive
definite:
\[
|c_i|<\sqrt{\frac{12 (2-\nu)}{(1-\nu)}}.
\]
We also have an upper bound on $|c_i|$ which guarantees that the
contact force is within its friction cone as follows:
\[
c_i^f=\frac{F_i^t}{F_i^n}=c_i \frac{2 (1-\nu)}{(2-\nu)}<\mu
\]
Finally, the third upper bound on $|c_i|$ is from the third
condition of Corollary~\ref{cor.grasp_stability}
\[
\|M^{-1/2} (K_p)_{as} M^{-1/2}\| < \sqrt{\alpha} \beta
\]
as discussed before. The selection of $|c_i|$s value is such that
it satisfies all these three upper bounds. This together with the
other two conditions, are sufficient for the equilibrium of the
grasping arrangement to be {\em locally asymptotically stable}.
 \epf

Based on Lemma~\ref{lemm.eqv_cond}, consider

Let $\Kg= \Kg_d^2 \Lg_{min}(\bar{M})/\norm{ \bar{M}}^2$, the  condition $\Gg \!<\! \sqrt{\alpha} \beta$ becomes

The asymmetry of $K_p$ strongly depends on the direction of the
contact forces, which in some cases can be selected during grasp
synthesis. The magnitude of the matrix norm of $(K_p)_{as}$
increases as the angle between the contact force and the normal at
the contacts increases.

